What is the answer to z=((i-2))/((1+2i))+((i-3))/((1+3i)) ?

The answer to z=((i-2))/((1+2i))+((i-3))/((1+3i)) is z=2i

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

z=((i-2))/((1+2i))+((i-3))/((1+3i))

z = \frac{( i - 2 )}{( 1 + 2 i )} + \frac{( i - 3 )}{( 1 + 3 i )}

z = 2 i

Solution steps

z = \frac{( i - 2 )}{( 1 + 2 i )} + \frac{( i - 3 )}{( 1 + 3 i )}

Simplify

\frac{i - 2}{1 + 2 i} + \frac{i - 3}{1 + 3 i} : 2 i

z = 2 i

6 x + 46 = x + 11

x, y = f (x - 8)

35 x = 11 mod 109

x, - 2 x - y = - 6

x, 5 y + 4 x - 2 = 0

What is the answer to z=((i-2))/((1+2i))+((i-3))/((1+3i)) ?
The answer to z=((i-2))/((1+2i))+((i-3))/((1+3i)) is z=2i